$g(t) = 3t$ $f(n) = n-6-3(g(n))$ $h(x) = 6x-2(g(x))$ $ g(f(1)) = {?} $
Solution: First, let's solve for the value of the inner function, $f(1)$ . Then we'll know what to plug into the outer function. $f(1) = 1-6-3(g(1))$ To solve for the value of $f$ , we need to solve for the value of $g(1)$ $g(1) = (3)(1)$ $g(1) = 3$ That means $f(1) = 1-6+(-3)(3)$ $f(1) = -14$ Now we know that $f(1) = -14$ . Let's solve for $g(f(1))$ , which is $g(-14)$ $g(-14) = (3)(-14)$ $g(-14) = -42$